# Why is $\int_{n}^{n+1}\int_{n}^x |f'(x)|\,dx \leq \int_{n}^{n+1} |f'(x)|\,dx$ for $x \in [n,n+1]$?

Can anyone show me how to prove the following:

For $x\in \left [ n,n+1 \right ]$: $$\int_{n}^{n+1}\int_{n}^{x}\left | f'(t) \right |dtdx\leqslant \int_{n}^{n+1}\left | f'(t)\right |dt?$$

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Since $n\leq x\leq n+1$: $\int_n^x|f'(t)|dt\leq \int_n^{n+1}|f'(t)|dt$ and if we integrate it with respect to $x$ over $[n,n+1]$, we get the expected result.

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@Davide Giraudo : Nice trick Davide. I really liked your proof. –  M.Krov Apr 8 '12 at 17:23

Hint: The integral runs over the domain $n\leq t\leq x\leq n+1$. Interchanging the $x$ and $t$ integration, we have $$\int_{n}^{n+1}\int_{n}^{x}\left | f^{'}(t) \right |dt\,dx = \int_{n}^{n+1}\int_{t}^{n+1}\left | f^{'}(t) \right |dx\,dt.$$ I am sure you can take it from here...

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@ Fabian: I can definitely take it from here. Thanks. –  M.Krov Apr 8 '12 at 17:24
$$\int_a^b \int_a^x g(t)\,dt\,dx=\int_a^b (b-x)g(x)\,dx$$
Why is this true? To go from the LHS to RHS intuitively, consider the fact that for $t\in [a,b]$, the double integral is set up to count $g(t)$ with generalized "multiplicity," or weight, proportionate to the length of the interval $[t,b]$ (this is the interval of $x$ values for which $t\in [a,x]$). To go from the RHS to the LHS algebraically, consider by-parts with $u=x-a,dv=g(x)$.